Conquer Club

Posted: **Thu Sep 27, 2007 9:00 am**

Did someone calculate to odds of winning in a single move?
It's easy to see that if there is one dice for each side - there's a 15/36 chance of winning.

What are the chances of winning when the dice are 3 against 2 (which is the most common)? What are the chances for a tie (each side loses one)?

Are there any known answers for that?

Posted: **Thu Sep 27, 2007 11:10 am**

Aerial Attack wrote:Here are the actual probabilities:

3 v 2 (A|D|T): 39.21%|28.63%|32.16%
7776 probable outcomes

2 v 2 (A|D|T): 22.76%|44.68%|32.56%
1296 probable outcomes

1 v 2 (A|D): 25.46%|74.54%
216 probable outcomes

3 v 1 (A|D): 65.97%|34.03%
1296 probable outcomes

2 v 1 (A|D): 57.87%|42.13%
216 probable outcomes

1 v 1 (A|D): 41.67%|58.33%
36 probable outcomes

Theories based on the numbers:

Always attack with 4+ armies (desperation?).
Always defend with 2+ armies (never leave any one alone?).

8 vs 10 is the first true 50/50 battle (you feeling lucky, punk?).
NOTE: I rounded 39.21 to 40 and 28.63 to 30 in order to have 10 die rolls work (in which case attacker loses 9 and defender loses 11!). Actually the first 8 rolls should result in 16 armies lost (7 attacker and 9 defender), meaning the last 2 rolls should be 2 v 1 [or 1 v 1] - at a loss of only 1 army per turn (resulting in either 1 attacker or 0 defender).

NOTE: Wouldn't 4 vs 5 work too? (No, because once attacker loses 1st army, they lose the 3rd attacking die).

Posted: **Fri Sep 28, 2007 7:19 am**

Aerial Attack wrote:
Aerial Attack wrote:8 vs 10 is the first true 50/50 battle (you feeling lucky, punk?).

This bit doesn't work. Use some Markov chains with those probabilities above, and you get these, assuming that the attacker continues to attack until he wins or has 1 army left, his odds of winning are:

50% with 6v5
39% with 6v6
29% with 6v7
63% with 7v5
52% with 7v6
.
.
35% with 8v10

Got Excel? Use these basic rules to figure your own odds, or modify them to figure new odds, say if you want to: stop if you get down to 3; stop if you have less than the defender; have a certain number left over; etc.

If you have 4 or more armies, say A, and your opponent has 2 or more , say B, then your odds of winning (P[A, B]) are 0.3921*P[A, B-2] + 0.2863*P[A-2, B] + 0.3216*P[A-1, B-1] . That is, you'll win AvB if you win two in the first roll and you can win AvB-2 or you lose two in the first roll and can win A-2vB or you win one, lose one and can win A-1vB-1. If you understood that, you can reckon the odds of winning with less than 3 or against 1 defender, then fire up a spreadsheet and generate odds for any size.

Posted: **Tue Oct 09, 2007 3:34 pm**

More on Markov Chains:
Markov Chains for the RISK Board Game Revisited - Jason A. Osborne

Posted: **Sun Oct 14, 2007 7:02 am**

Exact probabilities are less important than assessing one's risk appetite and being prepared to reevaluate an attack according to how it is going - so as soon a syou loose some armies you need to revealuate your probabilities from that point on - Personally I use experience and feel and even if the probabilities are wildly in one's favour, one still has to assess the almost inevitable army losses against the inevitable weakening of one's own armies - However, from all the comments below the most important to bear in mind is that when you are getting over run and trying to survive, spread your armies into 2 per country to make it harder to be eliminated and try to get some cards to change your position

Conquer Club

What are the odds

What are the odds

Probabilities